poj2187 旋转卡(qia)壳(ke)

题意：求凸包的直径

 

关于对踵点对、旋转卡壳算法的介绍可以参考这里：

http://www.cnblogs.com/Booble/archive/2011/04/03/2004865.html

http://www.cppblog.com/staryjy/archive/2009/11/19/101412.html

http://blog.csdn.net/ACMaker

 

这里使用了lrj的complex<double>大法来表示复数。

注意别忘了复数乘法的定义：(a+bi)*(c+di)=(ac-bd)+(bc+ad)i

 

#include <iostream>
#include <complex>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef complex<double> Point;      //Point A:complex   x+yi
typedef Point Vector;
const double eps=1e-10;

int dcmp(double x)      //return  0:x==0  -1:x<0  1:x>0
{
    if (fabs(x)<eps)    return 0;
        else return x<0?-1:1;
}

bool operator == (const Point &A,const Point &B)
{
    double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
    return dcmp(ax-bx)==0 && dcmp(ay-by)==0;
}
/*
bool operator <  (const Point &A,const Point &B)
{
    double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
    return ((ax<bx)||(ax==bx && ay<by));
}
*/
bool cmp(const Point &A,const Point &B)
{
    double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
    //return ((ax<bx)||(ax==bx && ay<by));
    int dx=dcmp(ax-bx),dy=dcmp(ay-by);   //return  0:ax==bx  -1:ax<bx  1:ax>bx
    return ((dx==-1)||((dx==0)&&(dy==-1)));
}


double Dot(Vector A,Vector B)        //Ax*Bx+Ay*By
{
    return real(conj(A)*B);
}

double Cross(Vector A,Vector B)     //Ax*By-Ay*Bx
{
    return imag(conj(A)*B);
}

Vector Rotate(Vector A,double rad)
{
    return A*exp(Point(0,rad));
}

double Dist(Point A,Point B)        //distance^2
{
    double ax=real(A),ay=imag(A),bx=real(B),by=imag(B);
    //cout<<ax<<" "<<ay<<" "<<bx<<" "<<by<<" "<<(ax-bx)*(ax-bx)+(ay-by)*(ay-by)<<endl;
    return ((ax-bx)*(ax-bx)+(ay-by)*(ay-by));
}

double PolygonArea(Point *p,int n)
{
    double area=0;
    for (int i=1;i<n-1;i++)
        area+=Cross(p[i]-p[0],p[i+1]-p[0]);
    return area/2;
}

int convexhull(Point *p,int n,Point *ch)
{
    sort(p,p+n,cmp);
    int m=0;
    for (int i=0;i<n;i++)
    {
        while (m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
            m--;
        ch[m++]=p[i];
    }
    int k=m;
    for (int i=n-2;i>=0;i--)
    {
        while (m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
            m--;
        ch[m++]=p[i];
    }
    if (n>1)    m--;
    return m;
}

double rotating_calipers(Point *ch,int num)
{
    int q=1;
    double ans=0;
    ch[num]=ch[0];
    for(int p=0;p<num;p++)
    {
        while(Cross(ch[p+1]-ch[p],ch[q+1]-ch[p])>Cross(ch[p+1]-ch[p],ch[q]-ch[p]))
            q=(q+1)%num;
        ans=max(ans,max(Dist(ch[p],ch[q]),Dist(ch[p+1],ch[q+1])));
    }
    return ans;
}

Point p[51000],ch[51000];
int n,x,y;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("ou.txt","w",stdout);

    while (cin>>n)
    {
        for (int i=0;i<n;i++)
        {
            cin>>x>>y;
            p[i]=Point(x,y);
        }

        int num=convexhull(p,n,ch);
        int ans=rotating_calipers(ch,num);

        //cout<<num<<" "<<ans<<endl;

        cout<<ans<<endl;
    }
    return 0;
}

 

发现一个下载USACO Contest数据的好地方：http://iskren.info/tasks/USACO/